# Pressure force in water on truncated cone with air inside

## Homework Statement

Consider the truncated cone tank submerged in water: inside the truncated cone tank there is air. Evaluate the forces acting on the truncated cone tank. ## The Attempt at a Solution

The forces are the following
• Boyuant force : $$F_b= \rho_w g V_{tank}$$
• Weight $$P=m g$$
• Normal reaction $$N$$
• Pressure force
The problem is with pressure force. In the solution of the exercise such force is calculated as $$F_p=\rho_W g L \pi R_1^2$$ I do not understand the reason of that ##R_1##. The water is distributed along the tank and, for istance, the upper face (of radius ##R_2##) is at a different pressure because it's higher.

I would have calculated it as follows.
$$F_p= F_{p \, on \, upper \, face}+F_{p \, on \, sides \, , y \, component}$$With ##F_{p \, on \, upper \, face}=\rho_W g (L-h) \pi R_2^2## and ##F_{p \, on \, sides \, , y \, component}= \int p(y) dS## (the last integral is for a force on a curved submerged surface so it is not very easy to calculate)

So what is the correct expression for the pressure force on the tank?

It is wrong. Fund of pressure force is Acchimede force or Boyuant force. Acchimede force is creaced by the difference between the up and down of object
So you dont need to find pressure force

haruspex
Homework Helper
Gold Member
It is wrong. Fund of pressure force is Acchimede force or Boyuant force. Acchimede force is creaced by the difference between the up and down of object
So you dont need to find pressure force
That is not in general true, and is not true here. Archimedes' principle only works when the liquid completely surrounds the object. In the present case it does not exert any pressure on the underside, so the net force of the water on the tank is the usual buoyancy force minus the force that it would exert on the underside if it were to completely surround the tank.
On the other hand, there is air in the tank, so instead of a pressure ρwgL+patmos acting on the underside we have a pressure pair. Unfortunately, we are not told what that pressure is. If the tank is making an airtight seal with the base it could be anything up to ρwgL+patmos.

@Soren4, your integral is correct and should produce the same answer, but as you can see, there is an easier way.

• Soren4
That is not in general true, and is not true here. Archimedes' principle only works when the liquid completely surrounds the object. In the present case it does not exert any pressure on the underside, so the net force of the water on the tank is the usual buoyancy force minus the force that it would exert on the underside if it were to completely surround the tank.
On the other hand, there is air in the tank, so instead of a pressure ρwgL+patmos acting on the underside we have a pressure pair. Unfortunately, we are not told what that pressure is. If the tank is making an airtight seal with the base it could be anything up to ρwgL+patmos.

Thanks for the answer! Yes it acts as a suction cup. But there surely is a force exerted on the tank because of the water above it and that's the pressure force I would like to know. Is the expression ##F_p= \rho_W g L \pi R_1^2## correct in this case?

haruspex
Homework Helper
Gold Member
Thanks for the answer! Yes it acts as a suction cup. But there surely is a force exerted on the tank because of the water above it and that's the pressure force I would like to know. Is the expression ##F_p= \rho_W g L \pi R_1^2## correct in this case?
No, that's not it.
Start with this: suppose it were not acting as a suction cup, i.e. the tank is a closed container and the water is all around it (though in an arbitrarily thin layer underneath it). What would the net force of the water on the tank be?

• Soren4
No, that's not it.
Start with this: suppose it were not acting as a suction cup, i.e. the tank is a closed container and the water is all around it (though in an arbitrarily thin layer underneath it). What would the net force of the water on the tank be?

Ok! Well in that case the pressure would act above, on the sides and below the tank and the resulting force would be the bouyant force I guess

The problem here is that the pressure is acting only above and on the sides and I cannot see the way to calculate the force..

haruspex
Homework Helper
Gold Member
Ok! Well in that case the pressure would act above, on the sides and below the tank and the resulting force would be the bouyant force I guess

The problem here is that the pressure is acting only above and on the sides and I cannot see the way to calculate the force..
Suppose the total force from the water is Fw (i.e. acting on top and sides), and the force that would act from the water on the base of the tank, if there were water there, is Fup.
Suppose further that the total force from the water on the tank, if there were water all around the tank, would be Fnet. What equation connects those three? How do any of them relate to Archimedes' "weight of water displaced" buoyancy force?

• Soren4
Suppose the total force from the water is Fw (i.e. acting on top and sides), and the force that would act from the water on the base of the tank, if there were water there, is Fup.
Suppose further that the total force from the water on the tank, if there were water all around the tank, would be Fnet. What equation connects those three? How do any of them relate to Archimedes' "weight of water displaced" buoyancy force?

Thanks a lot for your help! I would say that the relation is ##\vec{F}_{Archimedes}= \vec{F}_{net}= \vec{F}_{up} + \vec{F}_W##, so, if this is right, then I can get what I'm looking for, ##\vec{F}_W##, as ##\vec{F}_W=\vec{F}_{net}-\vec{F}_{up}=\rho_W V g - \rho_W g L \pi R_1^2##.

But in the problem ##\vec{F}_{Archimedes}## and ##\vec{F}_W## are considered completely separetly from each other so I'm not sure this would be correct..

haruspex
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